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\author{陈新亮}
\title{微分几何定理的证明}

\newtheorem{theorem}{定理}
\newtheorem{definition}{定义}

\begin{document}
\maketitle


\section{微分联络}

\begin{definition}[导数算符]
    映射 \(\nabla: \mathscr{F}_M(k, l) \rightarrow \mathscr{F}_M(k, l+1) \)
    称为 \(M\) 上的\emph{无挠导数算符}，如果它满足

    (1) 线性
    \begin{align}
        & \nabla_a(\alpha {T^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l} + \beta {S^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l}) \\
        &= \alpha \nabla_a {T^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l} + \beta \nabla_a {S^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l} \\
        & \forall {T^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l}, {S^{b_1, \cdots, b_k}}_{c_1, \cdots, c_l} \in \mathscr{F}_M(k, l)
    \end{align}

    (2) Leibnitz 律

    (3) 与缩并可交换

    (4) \(\nu(f) = \nu^a \nabla_a f \)

    (5) \(\nabla_a \nabla_b f = \nabla_b \nabla_a f\)

\end{definition}

\begin{theorem}[导数算符在标量场上的作用就是对偶矢量]
    设 \(\nabla_a\) 是任一导数算符, 则有
    \begin{equation}
        \nabla_a f = (df)_a , \qquad \forall f \in \mathscr{F}_M
    \end{equation}
    \begin{proof}
        \begin{align}
            & \forall \nu^a \in V = \mathcal{F}_M(1, 0) \\
            & (df)_a \nu^a = (df)_a \nu^\mu (\partial_\mu)^a
        \end{align}
    \end{proof}
\end{theorem}

\begin{theorem}[导数算符的局部性：只依赖于邻域]
    设 \(T_1, T_2 \in \mathscr{F}_M(k, l)\) 在 \(p \in M \) 的某邻域 \(N\) 内相等，即
    \( T_1\vert_N = T_2 \vert_N \),  则 \( \nabla_a T_1 \vert_p = \nabla_a T_2 \vert_p\) 。
\begin{proof}
    令 \( T = T_1 - T_2 \), 则 \(T\vert_N = 0\)。 
    取 \(S \in \mathscr{F}_M(k', l')\) 满足 \(S\vert_p \neq 0\) 且 \(S\vert_{M - N} = 0\)。
    于是有 \( TS = 0\)
    \begin{align}
        \nabla_a(TS)\vert_p = T\vert_p \nabla_a S\vert_p + S\vert_p \nabla_a T\vert_p = 0
    \end{align}
    将 \(T\vert_p = 0, S\vert_p \neq 0\) 代入可知
    \(\nabla_a T\vert_p = 0\) ，
    即 \(\nabla_a T_1\vert_p = \nabla_a T_2\vert_p\) 。
\end{proof}
\end{theorem}

\begin{theorem}[导数算符在对偶矢量上的作用只依赖于该点，而不需要邻域]
    设 $p\in M$ , $\omega_b, \omega_b^{\prime} \in \mathscr{F}_M(0, 1)$ 满足 $\omega_b\vert_p = \omega_b^{\prime}\vert_p$, 则
    \begin{equation}
        [(\tilde\nabla_a - \nabla_a)\omega_b^{\prime}]_p
        = [(\tilde\nabla_a - \nabla_a)\omega_b]_p
    \end{equation}
\begin{proof}[证明]
    只需证明
    \begin{equation}
        [\nabla_a (\omega_b^{\prime} - \omega_b)]_p
        = [\tilde\nabla_a(\omega_b^{\prime}  - \omega_b)]_p
    \end{equation}

    令 $\Omega_b = \omega_b^{\prime} - \omega_b \in \mathscr{F}_M(0, 1)$，则有 \(\Omega_b\vert_p = 0\)。

    设 $\{x^\mu\}$ 是 $p\in M$ 处的一个坐标系， 将 $\Omega_b$ 在该坐标系的基地上展开得

    \begin{equation}
        \Omega_b = \Omega_{\mu} (dx^\mu)_b
    \end{equation}
    其中 \(\Omega_{\mu} = \Omega_b \partial_\mu\) 满足 \(\Omega_{\mu}(p) = 0\)。

    因此
    \begin{align}
        [\nabla_a\Omega_b]_p &= [\nabla_a\Omega_\mu(dx^\mu)_b]_p \\
        &= \Omega_\mu(p) [\nabla_a (dx^\mu)]_p + [(dx^\mu)_b\nabla_a\Omega_\mu]_p \\
        &= [(dx^\mu)_b\nabla_a\Omega_\mu]_p
    \end{align}

    同理
    \begin{align}
        [\tilde\nabla_a\Omega_b]_p
        &= [(dx^\mu)_b\tilde\nabla_a\Omega_\mu]_p
    \end{align}

    由
    \begin{align*}
        [\tilde\nabla_a\Omega_\mu]_p = [\nabla_a\Omega_\mu]_p
    \end{align*}

    可知定理成立。
\end{proof}
\end{theorem}

由上述定理可知， 任意两个导数算符之差把对偶矢量线性地映射为另一个对偶矢量， 构成一个(1，2)型张量场。

\begin{align*}
    (\nabla_a - \tilde{\nabla_a})\omega_b = {C^c}_{ab}\omega_c
\end{align*}

\begin{theorem}[导数算符直接的关系]
    \begin{align*}
        \nabla_a \omega_b = \tilde{\omega_a}\omega_b - {C^c}_{ab}\omega_c \in \mathscr{F}_M(0, 1)
    \end{align*}
\end{theorem}

\begin{theorem}[无挠导数算符的对称性]
    \begin{align*}
        {C^c}_{ab} = {C^c}_{ba}
    \end{align*}
\end{theorem}



\end{document}
